this is a nice problem! it can be solved geometrically without using methods of multivariable calculus. here's my answer, which i think is correct:

Spoiler:

i believe the minimum distance is which is attained at two points: the reason is that the lines and are tangent to the circle

therefore implies that and thus

I was looking for such a nice solution, thanks NonCommAlg.
I guess you wonder if I have a solution or not?
Here it follows.

Spoiler:

Rotating a surface along a point does not changes the distance of the surface to that point.
Therefore, rotating with radians along , we have by making the substitution and .
Clearly is a hyperbola of two parts (I made this term up, and if possible teach me the correct one).
Hence, the distance attains it smallest value at the peak points , it is clear that .

I was looking for such a nice solution, thanks NonCommAlg.
I guess you wonder if I have a solution or not?
Here it follows.

Spoiler:

Rotating a surface along a point does not changes the distance of the surface to that point.
Therefore, rotating with radians along , we have by making the substitution and .
Clearly is a hyperbola of two parts (I made this term up, and if possible teach me the correct one).
Hence, the distance attains it smallest value at the peak points , it is clear that .

is actually a cylinder in cooridinates, along axis, with hyperbolic base . by the way, the Lagrange multipliers method will also solve the problem easily.